Let f a defined function in $\Bbb R+$ such as: $$f(x)=\sqrt{x\arctan{x}}$$
Prove that $\qquad $ $f^{-1}(x)=f(x) \leftrightarrow f(x)=x$
As an attempt, I've proven that f is a bijective function from $\Bbb R+$ to $\Bbb R+$, and I've proceeded by some equivalences : $$f(x)=x$$ $ \qquad \qquad \qquad \qquad \qquad \qquad \; \; \,\leftrightarrow f^{-1}(f(x))=f^{-1}(x)$ $ \qquad \qquad \qquad \qquad \qquad \qquad \; \; \,\leftrightarrow \qquad \quad \; \; x=f^{-1}(x)$ $ \qquad \qquad \qquad \qquad \qquad \qquad \; \; \,\leftrightarrow \qquad \; f(x)=f^{-1}(x)$
Is this true? Or is there any other method ?